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JeetGulia: Senior | Next Rank: 100 Posts; Posts: 66; Joined: Mon Jun 14, 2010 4:46 am; Thanked: 3 times

by JeetGulia » Tue Sep 14, 2010 3:19 am

A certain stock exchange designates each stock with one, two or three letter codes, where each letter is selected from the 26 letters. If the letters may be repeated and if the same letters used in a different order is a different code, how many different stocks codes are possibile.

a) 2,951
b) 8125
c) 15,600
d) 16,302
e) 18,278

ans E

Please help

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Source: Beat The GMAT — Problem Solving | Tue Sep 14, 2010 3:19 am

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Tue Sep 14, 2010 3:53 am

One letter code can be designated in 26 ways.
Two letter codes can be designated in 26^2 ways.
Three letter codes can be designated in 26^3 ways.
Therefore, number of stock codes possible = 26 + 26^2 + 26^3 = 18,278

The correct answer is [spoiler](E)[/spoiler].

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rb90: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Mon Jul 12, 2010 12:14 am; Location: India; GMAT Score:710

by rb90 » Mon Sep 27, 2010 5:03 am

But why dont we have to use permutation also in this sum? Since the sum hints at arrangement ..please help,im confused.

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euro: Master | Next Rank: 500 Posts; Posts: 120; Joined: Thu Sep 02, 2010 8:56 am; Location: India; Thanked: 7 times; GMAT Score:730

by euro » Mon Sep 27, 2010 8:18 am

rb90 wrote:But why dont we have to use permutation also in this sum? Since the sum hints at arrangement ..please help,im confused.

We don't use permutation {nPr formula} because no restriction is imposed here...
"the letters may be repeated and if the same letters used in a different order is a different code"

In case of two letter codes, the first position can be filled by any of the 26 alphabets and the second position can also be filled by any of the 26 alphabets. The two positions can be filled in 26 x 26 = 26^2 ways.

Similarly, for three-letter codes, the three positions can be filled in 26^3 ways (since repetition is allowed).

If repetition is allowed, for 'n' distinct objects taken 'r' at a time, the number of arrangements possible is n^r.

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msbinu: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Sat May 22, 2010 11:36 pm; Thanked: 3 times

by msbinu » Mon Sep 27, 2010 8:19 am

rb90 wrote:But why dont we have to use permutation also in this sum? Since the sum hints at arrangement ..please help,im confused.

Hi ,
We can use permutation also to solve this.

1 letter code can be selected from 26 letters using 26P1 ways = 26!/25! = 26
2 letter code could be selected in 26P1 * 26 P 1 ways = 26 * 26 = 676
3 letter code could be selected in 26p1 * 26P1 * 26 P1 ways = 26 * 26 * 26 = 17576

Hence total = 18278 .

Hope this helps

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